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DanielleElmas [232]
3 years ago
13

Rank the objects below in order from smallest to largest, where 1 is the smallest and 5 is the largest. local group of galaxies

_____ solar system _____ milky way galaxy _____ universe _____ earth
Physics
1 answer:
sasho [114]3 years ago
4 0
1. earth
2. solar system
3. milky way galaxy
4. local group of galaxies
5. universe
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What power of an engine is required to pump 2450N of water per second from a well 50m deep to the surface
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The correct one is A
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At an amusement park, a 7.00 kg swimmer uses a water slide to enter the main pool. The swimmer starts at rest, slides without fr
telo118 [61]

Answer:

27.44 J

Explanation:

We can find the energy at the top of the slide by using the potential energy equation:

  • PE = mgh

At the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.

The swimmer's mass is given as 7.00 kg.

The acceleration due to gravity is 9.8 m/s².

The (vertical) height of the water slide is 0.40 m.

Substitute these values into the potential energy equation:

  • PE = (7.00)(9.8)(0.40)
  • PE = 27.44

Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.

Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.

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2 years ago
A complex sound’s quality depends on the mix of harmonics or overtones added to the ______________ frequency
Brilliant_brown [7]
Fundamental frequency is the "base frequency" upon which the sound is built.
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3 years ago
Why are movie scenes in outer space often unrealistic?
kifflom [539]

Answer:

see explanation

Explanation:

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8 0
2 years ago
Read 2 more answers
Two charges are located in the x-y plane. If q1=-4.55 nC and is located at x=0.00 m, y=0.680 m and the second charge has magnitu
Elden [556K]

Answer:

Ex= -23.8 N/C  Ey = 74.3 N/C

Explanation:

As the  electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1, as follows:

Due  to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:

E₁ = k*(4.55 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0   E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:

r₂² = 1²m² + (0.6)²m² = 1.36 m²

The magnitude of the electric field due to  q2 can be found as follows:

E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the  cosine of  θ, is just, by definition, the opposite  of x/r₂:

⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855

By the same token, sin θ can be obtained as follows:

sin θ = - (0.6 m / 1.17 m) = -0.513

⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)

⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y =  88.6 N/C + (-14.3 N/C) =74.3 N/C

7 0
3 years ago
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